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3 years ago

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Triangle L M N is cut by line segment O P. Line segment O P goes from side M L to side M N. The length of O L is 14, the length

of O M is 28, the length of M P is y, and the length of P N is 18.
Which value of y would make O P is parallel to L N?

16
24
32
36

2 answers:

lions [1.4K]3 years ago

8 0

Answer:

The value of y that would make O P parallel to L N = 36

Step-by-step explanation:

This is a question on similar triangles. Find attached the diagram obtained from the given information.

Given:

The length of O L = 14

the length of O M = 28

the length of M P = y

the length of P N = 18

Length MN = MP + PN = y + 18

Length ML = MO + OL = 28+14 = 42

For OP to be parallel to LN,

MO/ML = MP/PN

MO/ML = 28/42

MP/PN= y/(y+18)

28/42 = y/(y+18)

42y = 28(y+18)

42y = 28y + 18(28)

42y-28y = 504

14y = 504

y = 504/14 = 36

The value of y that would make O P parallel to L N = 36

Salsk061 [2.6K]3 years ago

5 0

Answer:

D-36

Step-by-step explanation:

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Complete Question

From a random sample of size 18, a researcher states that (11.1, 15.7) inches is a 90% confidence interval for mu, the mean length of bass caught in a small lake. A normal distribution was assumed. Using the 90% confidence interval obtain:

a. A point estimate of $\mu$ and its 90% margin of error.

b. A 95% confidence interval for $\mu$ .

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a

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b

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Step-by-step explanation:

From the question we are told that

The sample size is n = 18

The 90% confidence interval is (11.1, 15.7)

Generally the point estimate of $\mu$ is mathematically evaluated as

$\= x = \frac{11.1 + 15.7 }{2}$

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Generally the margin of error is mathematically evaluated as

$E = \frac{15.7 - 11.1}{2 }$

=> $E = 2.3$

From the question we are told the confidence level is 90% , hence the level of significance is

$\alpha = (100 - 90 ) \%$

=> $\alpha = 0.10$

Generally from the normal distribution table the critical value of $\frac{\alpha }{2}$ is

$Z_{\frac{\alpha }{2} } = 1.645$

Generally the equation for the lower limit of the confidence interval is

$\= x - Z_{\frac{\alpha }{2} } * \frac{s}{\sqrt{18} } = 11.1$

=> $13.4 - 0.3877 s = 11.1$

=> $s = 5.932$

From the question we are told the confidence level is 95% , hence the level of significance is

$\alpha = (100 - 95 ) \%$

=> $\alpha = 0.05$

Generally from the normal distribution table the critical value of $\frac{\alpha }{2}$ is

$Z_{\frac{\alpha }{2} } = 1.96$

Generally the margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} } * \frac{\sigma }{\sqrt{n} }$

=> $E = 1.96 * \frac{5.932}{\sqrt{18} }$

=> $E = 2.7$

Generally 95% confidence interval is mathematically represented as

$\= x -E < \mu < \=x +E$

=> $13.4 - 2.7 < \mu < 13.4 + 2.7$

=> $10.7 < \mu < 16.1$

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